What torque would have to be applied to a rod with a length of #7 m# and a mass of #9 kg# to change its horizontal spin by a frequency #15 Hz# over #9 s#?

1 Answer
Feb 15, 2017

Answer:

The torque is #=384.8Nm#

Explanation:

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

The moment of inertia of a rod is #I=1/12*mL^2#

#=1/12*9*7^2= 147/4 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(15)/9*2pi#

#=((10pi)/3) rads^(-2)#

So the torque is #tau=147/4*(10pi)/3 Nm=1470/12piNm=384.8Nm#